‌n∕2[2a+(n−1)6]=(n−2)180∘ and an+3n2−3n=3n(n−2)180∘ ∴‌‌ Given a+(n−1)6∘=219∘ ⇒a=225∘−6n∘ Putting value of a in (i) We get (225−6n2)+3n2−3n=180n−360∘ ⇒2n2−42n−360=0 ⇒n2−14n−120=0 ⇒‌‌(n−20)(n+6)=0 ⇒n=20,−6 (Rejected) ∴‌‌n=20